height(s)=1/2(u+v)t 200=1/2(0+v)*1
v=400ft/sec
u = initial speed
v=final speed
Let v be the velocity when the ball is at 640 feets going downwards v = 48 feet /sec let the velocity with which it reaches the ground be u then, u2=v2+2gh g = acc due t ogravity in feet/sq.sec h = 640 feet the time taken to reach the ground = time to return to 640ft + the time to fall from there Time taken to get to the ground is 8 seconds. Final velocity is 208 feet per second downwards
The height, in feet, above the ground at time t, H(t) = 40 + 32*t - 16*t2
If the engineer's eye is at ground level, then the distance to the point on the building underneath its highest point is 450/tan(22) ft. If the engineer was standing and his eyes were x ft above the ground, the distance is (450-x)/tan(22) ft.
At any time 't' seconds after the ball is released,until it hits the ground,h = 5 + 48 t - 16.1 t2
8.5 + 2.75 = 11.25
31 m/s
mgh = 1/2 * m * v^2 v = sqrt (2 * 9.8 * 1.2) v = 4.8 m/s [down]
529.2 J
A 0.650 kg basketball is dropped out of a window that is 6.46 m above the ground. The ball is caught by a person whose hands are 1.32 m above the ground. How much work is done on the ball by its weight?
initial velocity, angle of launch, height above ground When a projectile is launched you can calculate how far it travels horizontally if you know the height above ground it was launched from, initial velocity and the angle it was launched at. 1) Determine how long it will be in the air based on how far it has to fall (this is why you need the height above ground). 2) Use your initial velocity to determine the horizontal component of velocity 3) distance travelled horizontally = time in air (part 1) x horizontal velocity (part 2)
initial velocity, angle of launch, height above ground When a projectile is launched you can calculate how far it travels horizontally if you know the height above ground it was launched from, initial velocity and the angle it was launched at. 1) Determine how long it will be in the air based on how far it has to fall (this is why you need the height above ground). 2) Use your initial velocity to determine the horizontal component of velocity 3) distance travelled horizontally = time in air (part 1) x horizontal velocity (part 2)
If it's fired horizontally, then its initial vertical velocity is zero. After that, the vertical velocityincreases by 9.8 meters per second every second, directed downward, and the projectile hitsthe ground after roughly 3.8 seconds.Exactly the same vertical motion as if it were dropped from the gun muzzle, with no horizontal velocity.
No because you touch yourself at night.|_ 4 VV |_Just kidding. It does!Psyche!Actually, it does not because when your mother dropped you from 2 meters above the ground, you fell at a negative VELOCITY. Speed does not specify direction, and therefore can not be negative.
1.56 seconds
My contractor suggested I invest in an above ground pool deck rather than the usual. What is the benefit?
lofty means high, so a lofty building is perhaps a building built high above the ground or maybe a skyscraper?
Assume that acceleration due to gravity, g = 9.8 ms-2. v2 = u2 + 2gs u, the initial velocity is 0 ms-1, s, the distance travelled is 140 - 20 = 120 m. So v2 = 2352 m2 so that v = 48.5 ms-1 approx.